What is the tangent of f(x)= 4/(x^2),\at x=1 ?

The tangent of f(x)= 4/(x^2),\at x=1 is y=-8x+12

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tangent of f(x)= 4/(x^2),\at x=1

tangent of

f (x) = \frac{4}{x ^{2}}, at x = 1

y = - 8 x + 12

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

f (x) = \frac{4}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{8}{x ^{3}}

Compute the slope of

f (x) = \frac{4}{x ^{2}}

x = 1 : m = - 8

Find the line with slope m=

- 8

and passing through

(1, 4) : y = - 8 x + 12

y = - 8 x + 12

\frac{d}{d x} (\frac{5 x ^{3} + 7 x ^{2}}{x})

\frac{d}{d x} (x - \frac{1}{9} x)

x \to 0 lim (e)

n = 1 \sum \infty \frac{n - 1}{3 n - 1}

y^{^{''}} + 1500 y = 320

What is the tangent of f(x)= 4/(x^2),\at x=1 ?
The tangent of f(x)= 4/(x^2),\at x=1 is y=-8x+12